# Introduction to Finite Fields (cont.)

Size: px
Start display at page:

Transcription

1 Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number p and contains a subfield isomorphic to Z p. Corollary (Isomorphism to Z p ). Any field F with prime order p is isomorphic to Z p. Theorem (Prime Power Fields). There exists a field F of order n. n is a power of a prime number. 6.2 Polynomials over a field Definition 6.1. Let F be a field. The set n F [x] := { a i x i a i F, n Z 0 } i=0 is called the polynomial ring over F. An element of F [x] is called a polynomial over F. For a polynomial f(x) = n a i x i, if a n 0 then the integer n is called the degree of f(x), i=0 denoted by deg(f(x)), (for convenience, we define deg(0) = ). Furthermore, a nonzero polynomial f(x) = n a i x i of degree n is said to be monic if a n = 1. A polynomial f(x) i=0 59

2 60 Math 422. Coding Theory of positive degree is said to be reducible (over F ) if there exist two polynomials g(x) and h(x) over F such that deg(g(x)) < deg(f(x)), deg(h(x)) < deg(f(x)) and f(x) = g(x)h(x). Otherwise, the polynomial f(x) of positive degree is said to be irreducible (over F ). Example ). The polynomial f(x) = x 4 +2x 6 Z 3 [x] is of degree 6. It is reducible as f(x) = x 4 (1 + 2x 2 ). 2). The polynomial g(x) = 1+x+x 2 Z 2 [x] is of degree 2. It is irreducible. Otherwise, it would have a linear factor x or x + 1; i.e., 0 or 1 would be a root of g(x), but g(0) = g(1) = 1 Z 2. 3). Using the same arguments as in 2), we can show that both 1+x+x 3 and 1+x 2 +x 3 are irreducible over Z 2 as they have no linear factors. Definition 6.2 (Division Rule). Let f(x) F [x] be a polynomial of degree n 1. Then, for any polynomial g(x) F [x], there exists a unique pair (s(x), r(x)) of polynomials with deg(r(x)) < deg(f(x)) or r(x) = 0 such that g(x) = s(x)f(x)+r(x). The polynomial r(x) is called the (principal) remainder of g(x) divided by f(x). If r(x) = 0, then we call f(x) divides g(x) or f(x) is a divisor of g(x) and g(x) is divisible by f(x) g(x) is called a multiple of f(x). Example 6.2. For example, let f(x) = 1 + x 2 and g(x) = x + 2x 4 be two polynomials in Z 5 [x]. Since we have g(x) = x+2x 4 = (3+2x 2 )(1+x 2 )+(2+x) = (3+2x 2 )f(x)+(2+x), the remainder of g(x) divided by f(x) is 2 + x. Lecture 18, March 17, 2011 Lemma 6.3 (Linear Factors). A polynomial f(x) has a linear factor x a. f(a) = 0. Proof. By the division rule, f(x) has a linear factor x a. f(x) = (x a)g(x), i.e., the principal reminder is 0. f(a) = 0. Corollary 6.4 (Irreducible 2nd or 3rd Degree Polynomials). A polynomial f(x) in F [x] of degree 2 or 3 is irreducible. f(a) 0 for all a F. Proof. f(x) can be factored into polynomials of smaller degree. it has at least one linear factor (x a). f(a) = 0, by Lemma. Example ). The polynomial g(x) = 1 + x + x 2 Z 2 [x] is of degree 2. It is irreducible. Since g(0) = g(1) = 1 Z 2.

3 6.2. Polynomials over a field 61 2). Using the corollary, we can show that both 1 + x + x 3 and 1 + x 2 + x 3 are irreducible over Z 2. Analogous to the integral ring Z, we can introduce the following notions. Definition 6.5. Let f(x), g(x) F [x] be two nonzero polynomials. The greatest common divisor of f(x), g(x), denoted by gcd(f(x), g(x)), is the monic polynomial of the highest degree which is a divisor of both f(x) and g(x). In particular, we say that f(x) is coprime (or prime) to g(x) if gcd(f(x), g(x)) = 1. The least common multiple of f(x), g(x), denoted by lcm(f(x), g(x)), is the monic polynomial of the lowest degree which is a multiple of both f(x) and g(x). Remark. 1). If f(x) and g(x) have the following factorizations: f(x) = ap 1 (x) e 1 p 2 (x) e2 p n (x) en, g(x) = bp 1 (x) d1 p n (x) dn, where a, b F \ {0}, e i, d i 0 and p i (x) are distinct monic irreducible polynomials, then gcd(f(x), g(x)) = p 1 (x) min{e 1,d 1} p n (x) min{en,dn} and lcm(f(x), g(x)) = p 1 (x) max{e 1,d 1} p n (x) max{en,dn}. 2). Let f(x), g(x) F [x] be two nonzero polynomials. Then there exist two polynomials u(x), v(x) with deg(u(x)) < deg(g(x)) and deg(v(x)) < deg(f(x)) such that gcd(f(x), g(x)) = u(x)f(x) + v(x)g(x). 3). It is easily shown that gcd(f(x)h(x), g(x)) = gcd(f(x), g(x)) if gcd(h(x), g(x)) = 1. Definition 6.6. Let f(x) be a fixed polynomial in F [x]. Two polynomials g(x) and h(x) in F [x] is said to be congruent modulo f(x), symbolized by g(x) h(x)(mod f(x)) if g(x) h(x) is divisible by f(x). By the division rule, any polynomial g(x) in F [x] is congruent module f(x) to a unique polynomial r(x) of degree less than deg(f(x)) (r(x) is just the principal remainder when g(x) is divided by f(x)). We denote F [x]/(f(x)) or F [x]/f(x) the set of polynomials in F [x] of degree less than deg(f(x)), with addition and multiplication carried out modulo f(x) as follows:

4 62 Math 422. Coding Theory Suppose g(x) and h(x) belong to F [x]/f(x). Then the sum g(x) + h(x) in F [x]/f(x) is the same as the sum in F [x], because deg(g(x) + h(x)) < deg(f(x)). The product g(x)h(x) is the principal remainder when g(x)h(x) is divided by f(x). There are many analogies between the integral ring Z and a polynomial ring F [x]. Theorem 6.1. Let f(x) be a polynomial over a field F of degree 1. Then F [x]/(f(x)), together with the addition and multiplication, forms a ring. Furthermore, F [x]/(f(x)) is a field if and only if f(x) is irreducible. Proof. It is easy to verify that F [x]/(f(x)) is a ring. By applying exactly the same arguments as in the proof of Theorem 1.5, we can prove the second part. Example ). Consider the ring R[x]/(1 + x 2 ) = {a + bx a, b R}. It is a field since 1 + x 2 is irreducible over R. In fact, it is isomorphic to the complex field C! To see this, we just replace x in R[x]/(1 + x 2 ) by the imaginary unit i. 2). Consider the ring Z 2 [x]/(1 + x 2 ) = {0, 1, x, 1 + x}. We construct the addition and multiplication tables as follows. We see from the multiplication table that Z 2 [x]/(1 + x 2 ) is not a field as (1 + x)(1 + x) = 0. 3). Consider the ring Z 2 [x]/(1 + x + x 2 ) = {0, 1, x, 1 + x}. As 1 + x + x 2 is irreducible over Z 2, the ring Z 2 [x]/(1 + x + x 2 ) is in fact a field. This can also be verified by the addition and multiplication tables. Remark. If F = F q is a finite field, then for polynomial f(x) with degree n we have F [x]/(f(x)) = q n. Indeed, the ring F [x]/f(x) consists of all polynomials of degree n 1. Each of the n coefficients of such a polynomial belongs to F q, i.e., has q choices. 6.3 Structure of finite fields We list some facts about finite fields in this section: Lemma 6.7. For every element α of a finite field F q, we have α q = α. Corollary 6.8. Let F be a subfield of E with F = q. Then an element α of E lies in F if and only if α q = α. Theorem 6.2. For any prime p and integer n 1, there exists a unique finite field of p n elements.

5 6.3. Structure of finite fields 63 Lecture 19, March 22, 2011 Recall. For an irreducible polynomial f(x) of degree n over a field F, let α be a root of f(x) = 0. Then the field F [x]/(f(x)) can be represented as F [α] = {a 0 + a 1 α + + a n 1 α n 1 a i F } if we replace x in F [x]/(f(x)) by α. An advantage of using F [α] to replace the field F [x]/(f(x)) is that we can avoid the confusion between an element of F [x]/(f(x)) and a polynomial over F. Definition 6.9 (Primitive element). An element α in a finite field F q primitive element (or generator) of F q if F q = {0, α, α 2,..., α q 1 }. is called a Example 6.5. Consider the field F 4 = F 2 [x]/(x 2 +x+1) = F 2 [α] = {0, 1, x, x+1}, where α is a root of the irreducible polynomial 1 + x + x 2 F 2 [x]. Then we have α 2 = (1 + α) = 1 + α, α 3 = α(α 2 ) = α(1 + α) = α + α 2 = 1. Thus, F 4 = {0, α, α 2, α 3 }, so α is a primitive element. Definition 6.10 (Order). The order of a nonzero element α F q, denoted by ord(α), is the smallest positive integer k such that α k = 1. Example 6.6. Since there are no linear factors for the polynomial 1+x 2 over F 3, 1+x 2 is irreducible over F 3. Consider the element α in the field F 9 = F 3 [x]/(x 2 +1) = F 3 [α], where α is a root of 1 = x 2. Then α 2 = 1, α 3 = α(α 2 ) = α and α 4 = (α 2 ) 2 = ( 1) 2 = 1. This means that ord(α) = 4. Lemma ). The order ord(α) divides q 1 for every α F q. 2). For two nonzero elements α, β F q, if gcd(ord(α), ord(β)) = 1, then ord(α β) = ord(α) ord(β). Theorem ). A nonzero element of F q is a primitive element if and only if its order is q 1. 2). Every finite field has at least one primitive element. Remark. 1). Primitive elements are not unique. 2). If α is a root of an irreducible polynomial f(x) of degree m over F q, and it is also a primitive element of F q m = F [x]/f(x) = F q [α], then every element in F q m can be represented both as a polynomial in α and as a power of α, since F q m = {a 0 + a 1 α + + a m 1 α m 1 a i F q } = {0, α, α 2,..., α qm 1 }.

6 64 Math 422. Coding Theory Addition for the elements of F q m is easily carried out if the elements are represented as polynomials in α, whilst multiplication is easily done if the elements are represented as powers of α. Example ). Let α be a root of 1 + x + x 3 F 2 [x]. Hence, F 8 = F 2 [α]. The order of α is a divisor of 8 1 = 7. Thus, ord(α) = 7 and α is a primitive element. In fact, any nonzero element in F 8 except 1 is a primitive element, since all the elements like this is not order of 1. 2). Let α be a root of 1 + 2x + x 3 F 3 [x]. This polynomial is irreducible over F 3 as it has no linear factors. Hence, F 27 = F 3 [α]. The order of α is a divisor of 27 1 = 26. Thus, ord(α) is 2, 13 or 26. First, ord(α) 2; otherwise, α would be 1 or 1, neither of which is a root of 1 + 2x + x 3. Furthermore, we have α 13 = 1 1, indeed α 13 = α α 3 (α 3 ) 3 = α ( 2α 1) ( 2α 1) 3 = α ( 2α 1) ( 8α 3 1) = α ( 2α 1) (α 3 1) = α (α 1) (α 2) = α 3 + 2α = 1 Thus, ord(α) = 26 and α is a primitive element of F Minimal polynomials Definition 6.12 (Minimal polynomial). A minimal polynomial of an element α F q m with respect to F q is a nonzero monic polynomial f(x) of the least degree in F q [x] such that f(α) = 0. Example 6.8. Let α be a root of the polynomial 1 + x + x 2 F 2 [x]. It is clear that the two linear polynomials x and 1 + x are not minimal polynomials of α. Therefore, 1 + x + x 2 is a minimal polynomial of α. Since 1 + (1 + α) + (1 + α) 2 = α α 2 = 1 + α + α 2 = 0 and 1 + α is not a root of x or 1 + x, thus 1 + x + x 2 is also a minimal polynomial of 1 + α. Theorem 6.4. Let β F p r. If f(x) F p [x] has β as a root, then f(x) is divisible by the minimal polynomial of β. Corollary The minimal polynomial of an element of a field F q divides x q x. Theorem ). The minimal polynomial of an element of F q m with respect to F q exists and is unique. It is also irreducible over F q.

7 6.4. Minimal polynomials 65 2). If a monic irreducible polynomial M(x) F q [x] has α F q m as a root, then it is the minimal polynomial of α with respect to F q. Definition 6.14 (cyclotomic coset). Let r be co-prime to q. The cyclotomic coset of q (or q-cyclotomic coset) modulo r containing i is defined by C i = {(i q j (mod r)) Z r j = 0, 1,...}. A subset {i 1,..., i t } of Z r is called a complete set of representatives of cyclotomic cosets of q modulo r if C i1,..., C it are distinct and t j=1 C i j = Z r. Remark. 1). It is easy to verify that two cyclotomic cosets are either equal or disjoint. Hence, the cyclotomic cosets partition Z r. Lecture 20, March 24, ). If r = q m 1 for some m 1, each cyclotomic coset contains at most m elements, as q m 1(mod q m 1). 3). It is easy to see that, in the case of r = q m 1 for some m 1, if gcd(i, q m 1) = 1 then C i = m. Example ). Consider the cyclotomic cosets of 2 modulo 15: C 0 = {0}, C 1 = {1, 2, 4, 8}, C 3 = {3, 6, 9, 12}, C 5 = {5, 10}, C 7 = {7, 11, 13, 14}. Thus, C 1 = C 2 = C 4 = C 8, and so on. The set {0, 1, 3, 5, 7} is a complete set of representatives of cyclotomic cosets of 2 modulo 15. The set {0, 1, 6, 10, 7} is also a complete set of representatives of cyclotomic cosets of 2 modulo 15. 2). Consider the cyclotomic cosets of 3 modulo 26: C 0 = {0}, C 1 = {1, 3, 9}, C 2 = {2, 6, 18}, C 4 = {4, 12, 10}, C 5 = {5, 15, 19}, C 7 = {7, 21, 11}, C 8 = {8, 24, 20}, C 13 = {13}, C 14 = {14, 16, 22}, C 17 = {17, 25, 23}. In this case, we have C 1 = C 3 = C 9, and so on. The set {0, 1, 2, 4, 5, 7, 8, 13, 14, 17} is a complete set of representatives of cyclotomic cosets of 3 modulo 26. field. We are now ready to determine the minimal polynomials for all the elements in a finite

8 66 Math 422. Coding Theory Theorem 6.6. Let α be a primitive element of F q m. Then the minimal polynomial of α i is M (i) (x) = (x α j ), j C i where C i is the unique cyclotomic coset of q modulo q m 1 containing i. Remark. 1). The degree of the minimal polynomial of α i is equal to the size of the cyclotomic coset containing i. 2). From the Theorem, we know that α i and α k have the same minimal polynomial if and only if i, k are in the same cyclotomic coset. Example Let α be a root of 2 + x + x 2 F 3 [x]; i.e., 2 + α + α 2 = 0. Then the minimal polynomial of α as well as α 3 is 2 + x + x 2. The minimal polynomial of α 2 is M (2) (x) = (x α j ) = (x α 2 )(x α 6 ) = α 8 (α 2 + α 6 )x + x 2. j C 2 We know that α 8 = 1 as α F 9. To find M (2) (x), we have to simplify α 2 + α 6. We make use of the relationship above to obtain α 2 + α 6 = (1 α) + (1 α) 3 = 2 α α 3 = 2 α α(1 α) = 2 2α + α 2 = 0. Hence, the minimal polynomial of α 2 is 1 + x 2. In the same way, we may obtain the minimal polynomial 2 + 2x + x 2 of α 5. The following result will be useful when we study cyclic codes. Theorem 6.7. Let n be a positive integer with gcd(q, n) = 1. Suppose that m is a positive integer satisfying n (q m 1). Let α be a primitive element of F q m and let M (j) (x) be the minimal polynomial of α j with respect to F q. Let {s 1, s 2,..., s t } be a complete set of representatives of cyclotomic cosets of q modulo n. Then the polynomial x n 1 has the factorization into monic irreducible polynomials over F q : x n 1 = t i=1 M ( (qm 1)s i n ) (x). Remark. 1). In the theorem, the set {s 1, s 2,..., s t } is a complete set of representatives of cyclotomic cosets of q modulo n, but by Theorem 6.6 the minimal polynomial M ( (qm 1)s i n ) (x) is given through complete set of representatives of cyclotomic cosets of q modulo q m 1.

9 6.4. Minimal polynomials 67 2). The factorization of x n 1 into monic irreducible polynomials over F q doesn t depend on the choice of m. Corollary Let n be a positive integer with gcd(q, n) = 1. Then the number of monic irreducible factors of x n 1 over F q is equal to the number of cyclotomic cosets of q modulo n. Example ). Consider the polynomial x 13 1 over F 3. It is easy to check that {0, 1, 2, 4, 7} is a complete set of representatives of cyclotomic cosets of 3 modulo 13. Since 13 is a divisor of 3 3 1, we consider the field F 27. Let α be a root of 1 + 2x + x 3. By the Example, α is a primitive element of F 27. By Example, we know all the cyclotomic cosets of 3 modulo 26 containing multiples of 2. Hence, we obtain M (0) (x) = x + 2, M (2) (x) = j C 2 (x α j ) = (x α 2 )(x α 6 )(x α 18 ) = 2 + x + x 2 + x 3, M (4) (x) = j C 4 (x α j ) = (x α 4 )(x α 12 )(x α 10 ) = 2 + x 2 + x 3 M (8) (x) = j C 8 (x α j ) = (x α 8 )(x α 20 )(x α 24 ) = 2 + 2x + 2x 2 + x 3 M (14) (x) = j C 14 (x α j ) = (x α 14 )(x α 16 )(x α 22 ) = 2 + 2x + x 3. By the Theorem, we obtain the factorization of x 13 1 over F 3 into monic irreducible polynomials: x 13 1 = M (0) (x) M (2) (x) M (4) (x) M (8) (x) M (14) (x) = (x + 2)(2 + x + x 2 + x 3 )(2 + x 2 + x 3 )(2 + 2x + 2x 2 + x 3 )(2 + 2x + x 3 ). 2). We can also discuss the polynomial x 21 1 over F 2. It is easy to check that {0, 1, 3, 5, 7, 9} is a complete set of representatives of cyclotomic cosets of 2 modulo 21. Since 21 is a divisor of 2 6 1, we consider the field F 64. Let α be a root of 1 + x + x 6. It can be verified that α is a primitive element of F 64 (check that α 3 1, α 7 1, α 9 1 and α 21 1). We list the cyclotomic cosets of 2 modulo 63 containing multiples of 3: C 0 = {0}, C 3 = {36, 12, 24, 48, 33}, C 9 = {9, 18, 36}, C 15 = {15, 30, 60, 57, 51, 39}, C 21 = {21, 42}, C 27 = {27, 54, 45}.

10 68 Math 422. Coding Theory Hence, we obtain M (0) (x) = x + 1, M (3) (x) = j C 3 (x α j ) = 1 + x + x 2 + x 4 + x 6, M (9) (x) = j C 9 (x α j ) = 1 + x 2 + x 3, M (15) (x) = j C 15 (x α j ) = 1 + x 2 + x 4 + x 5 + x 6, M (21) (x) = j C 21 (x α j ) = 1 + x + x 2, M (27) (x) = j C 27 (x α j ) = 1 + x + x 3. By the above Theorem, we obtain the factorization of x 21 1 over F 2 into monic irreducible polynomials: x 21 1 = M (0) (x) M (3) (x) M (9) (x) M (15) (x) M (21) (x) M (27) (x) = (x + 1) (1 + x + x 2 + x 4 + x 6 ) (1 + x 2 + x 3 ) (1 + x 2 + x 4 + x 5 + x 6 ) (1 + x + x 2 ) (1 + x + x 3 ). Remark. We can find the factorization of x n 1 over F q step by step as following: Step 1: Give a complete set of representatives of cyclotomic cosets of q modulo n. Step 2: Find a positive integer m such that n q m 1. Step 3: Give a primitive element α of the field F q m. Step 4: List the corresponding cyclotomic cosets of q modulo q m 1 and using Theorem 6.6 to write down the minimal polynomials. Step 5: Get the factorization through the minimal polynomials.

### minimal polyonomial Example

Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We

### The Division Algorithm for Polynomials Handout Monday March 5, 2012

The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following,

### PROBLEM SET 6: POLYNOMIALS

PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other

### Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.

Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize

### CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY

January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.

### H/wk 13, Solutions to selected problems

H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots.

### Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013

Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of

### 7. Some irreducible polynomials

7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of

### Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)

Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 Remainder and Factor Theorem 15 Definition of factor If f (x)

### Factorization Algorithms for Polynomials over Finite Fields

Degree Project Factorization Algorithms for Polynomials over Finite Fields Sajid Hanif, Muhammad Imran 2011-05-03 Subject: Mathematics Level: Master Course code: 4MA11E Abstract Integer factorization is

### Quotient Rings and Field Extensions

Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.

### it is easy to see that α = a

21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore

### 1 = (a 0 + b 0 α) 2 + + (a m 1 + b m 1 α) 2. for certain elements a 0,..., a m 1, b 0,..., b m 1 of F. Multiplying out, we obtain

Notes on real-closed fields These notes develop the algebraic background needed to understand the model theory of real-closed fields. To understand these notes, a standard graduate course in algebra is

### ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS

ON GALOIS REALIZATIONS OF THE 2-COVERABLE SYMMETRIC AND ALTERNATING GROUPS DANIEL RABAYEV AND JACK SONN Abstract. Let f(x) be a monic polynomial in Z[x] with no rational roots but with roots in Q p for

### How To Prove The Dirichlet Unit Theorem

Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if

### calculating the result modulo 3, as follows: p(0) = 0 3 + 0 + 1 = 1 0,

Homework #02, due 1/27/10 = 9.4.1, 9.4.2, 9.4.5, 9.4.6, 9.4.7. Additional problems recommended for study: (9.4.3), 9.4.4, 9.4.9, 9.4.11, 9.4.13, (9.4.14), 9.4.17 9.4.1 Determine whether the following polynomials

### MOP 2007 Black Group Integer Polynomials Yufei Zhao. Integer Polynomials. June 29, 2007 Yufei Zhao yufeiz@mit.edu

Integer Polynomials June 9, 007 Yufei Zhao yufeiz@mit.edu We will use Z[x] to denote the ring of polynomials with integer coefficients. We begin by summarizing some of the common approaches used in dealing

### Factoring Polynomials over Finite Fields

Enver Ozdemir 1 F p, p is an odd prime. 2 f (x) F p [x] 3 The Problem: Find f i (x) F p [x], f (x) = f 1 (x)... f n (x), f i (x) irreducible and coprime. 1 F p, p is an odd prime. 2 f (x) F p [x] 3 The

### Math 319 Problem Set #3 Solution 21 February 2002

Math 319 Problem Set #3 Solution 21 February 2002 1. ( 2.1, problem 15) Find integers a 1, a 2, a 3, a 4, a 5 such that every integer x satisfies at least one of the congruences x a 1 (mod 2), x a 2 (mod

### r + s = i + j (q + t)n; 2 rs = ij (qj + ti)n + qtn.

Chapter 7 Introduction to finite fields This chapter provides an introduction to several kinds of abstract algebraic structures, particularly groups, fields, and polynomials. Our primary interest is in

This page intentionally left blank Coding Theory A First Course Coding theory is concerned with successfully transmitting data through a noisy channel and correcting errors in corrupted messages. It is

### I. GROUPS: BASIC DEFINITIONS AND EXAMPLES

I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called

### 3 1. Note that all cubes solve it; therefore, there are no more

Math 13 Problem set 5 Artin 11.4.7 Factor the following polynomials into irreducible factors in Q[x]: (a) x 3 3x (b) x 3 3x + (c) x 9 6x 6 + 9x 3 3 Solution: The first two polynomials are cubics, so if

### 9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11.

9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n-1 x n-1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role

### PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.

PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include

### = 2 + 1 2 2 = 3 4, Now assume that P (k) is true for some fixed k 2. This means that

Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without

### FACTORING SPARSE POLYNOMIALS

FACTORING SPARSE POLYNOMIALS Theorem 1 (Schinzel): Let r be a positive integer, and fix non-zero integers a 0,..., a r. Let F (x 1,..., x r ) = a r x r + + a 1 x 1 + a 0. Then there exist finite sets S

### Factoring of Prime Ideals in Extensions

Chapter 4 Factoring of Prime Ideals in Extensions 4. Lifting of Prime Ideals Recall the basic AKLB setup: A is a Dedekind domain with fraction field K, L is a finite, separable extension of K of degree

### Factoring Polynomials

Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent

### Winter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com

Polynomials Alexander Remorov alexanderrem@gmail.com Warm-up Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).

### Die ganzen zahlen hat Gott gemacht

Die ganzen zahlen hat Gott gemacht Polynomials with integer values B.Sury A quote attributed to the famous mathematician L.Kronecker is Die Ganzen Zahlen hat Gott gemacht, alles andere ist Menschenwerk.

### Prime Numbers and Irreducible Polynomials

Prime Numbers and Irreducible Polynomials M. Ram Murty The similarity between prime numbers and irreducible polynomials has been a dominant theme in the development of number theory and algebraic geometry.

### Chapter 1. Search for Good Linear Codes in the Class of Quasi-Cyclic and Related Codes

Chapter 1 Search for Good Linear Codes in the Class of Quasi-Cyclic and Related Codes Nuh Aydin and Tsvetan Asamov Department of Mathematics, Kenyon College Gambier, OH, USA 43022 {aydinn,asamovt}@kenyon.edu

### The cyclotomic polynomials

The cyclotomic polynomials Notes by G.J.O. Jameson 1. The definition and general results We use the notation e(t) = e 2πit. Note that e(n) = 1 for integers n, e(s + t) = e(s)e(t) for all s, t. e( 1 ) =

### Cyclotomic Extensions

Chapter 7 Cyclotomic Extensions A cyclotomic extension Q(ζ n ) of the rationals is formed by adjoining a primitive n th root of unity ζ n. In this chapter, we will find an integral basis and calculate

### FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z

FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization

### Unique Factorization

Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon

### (a) Write each of p and q as a polynomial in x with coefficients in Z[y, z]. deg(p) = 7 deg(q) = 9

Homework #01, due 1/20/10 = 9.1.2, 9.1.4, 9.1.6, 9.1.8, 9.2.3 Additional problems for study: 9.1.1, 9.1.3, 9.1.5, 9.1.13, 9.2.1, 9.2.2, 9.2.4, 9.2.5, 9.2.6, 9.3.2, 9.3.3 9.1.1 (This problem was not assigned

### The finite field with 2 elements The simplest finite field is

The finite field with 2 elements The simplest finite field is GF (2) = F 2 = {0, 1} = Z/2 It has addition and multiplication + and defined to be 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 0 0 = 0 0 1 = 0

### A NOTE ON FINITE FIELDS

A NOTE ON FINITE FIELDS FATEMEH Y. MOKARI The main goal of this note is to study finite fields and their Galois groups. Since I define finite fields as subfields of algebraic closure of prime fields of

### GROUPS ACTING ON A SET

GROUPS ACTING ON A SET MATH 435 SPRING 2012 NOTES FROM FEBRUARY 27TH, 2012 1. Left group actions Definition 1.1. Suppose that G is a group and S is a set. A left (group) action of G on S is a rule for

### EXERCISES FOR THE COURSE MATH 570, FALL 2010

EXERCISES FOR THE COURSE MATH 570, FALL 2010 EYAL Z. GOREN (1) Let G be a group and H Z(G) a subgroup such that G/H is cyclic. Prove that G is abelian. Conclude that every group of order p 2 (p a prime

### POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS

POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0).

### 10 Splitting Fields. 2. The splitting field for x 3 2 over Q is Q( 3 2,ω), where ω is a primitive third root of 1 in C. Thus, since ω = 1+ 3

10 Splitting Fields We have seen how to construct a field K F such that K contains a root α of a given (irreducible) polynomial p(x) F [x], namely K = F [x]/(p(x)). We can extendthe procedure to build

### Galois Theory III. 3.1. Splitting fields.

Galois Theory III. 3.1. Splitting fields. We know how to construct a field extension L of a given field K where a given irreducible polynomial P (X) K[X] has a root. We need a field extension of K where

### U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra

U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory

FINITE FIELDS KEITH CONRAD This handout discusses finite fields: how to construct them, properties of elements in a finite field, and relations between different finite fields. We write Z/(p) and F p interchangeably

### Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field

Math 345-60 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will

### QUADRATIC RECIPROCITY IN CHARACTERISTIC 2

QUADRATIC RECIPROCITY IN CHARACTERISTIC 2 KEITH CONRAD 1. Introduction Let F be a finite field. When F has odd characteristic, the quadratic reciprocity law in F[T ] (see [4, Section 3.2.2] or [5]) lets

### Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients

DOI: 10.2478/auom-2014-0007 An. Şt. Univ. Ovidius Constanţa Vol. 221),2014, 73 84 Irreducibility criteria for compositions and multiplicative convolutions of polynomials with integer coefficients Anca

### FACTORING AFTER DEDEKIND

FACTORING AFTER DEDEKIND KEITH CONRAD Let K be a number field and p be a prime number. When we factor (p) = po K into prime ideals, say (p) = p e 1 1 peg g, we refer to the data of the e i s, the exponents

### ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS

ABSTRACT ALGEBRA: A STUDY GUIDE FOR BEGINNERS John A. Beachy Northern Illinois University 2014 ii J.A.Beachy This is a supplement to Abstract Algebra, Third Edition by John A. Beachy and William D. Blair

### THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS

THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear

### Ideal Class Group and Units

Chapter 4 Ideal Class Group and Units We are now interested in understanding two aspects of ring of integers of number fields: how principal they are (that is, what is the proportion of principal ideals

### JUST THE MATHS UNIT NUMBER 1.8. ALGEBRA 8 (Polynomials) A.J.Hobson

JUST THE MATHS UNIT NUMBER 1.8 ALGEBRA 8 (Polynomials) by A.J.Hobson 1.8.1 The factor theorem 1.8.2 Application to quadratic and cubic expressions 1.8.3 Cubic equations 1.8.4 Long division of polynomials

### Mathematics Course 111: Algebra I Part IV: Vector Spaces

Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 1996-7 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are

### Integer roots of quadratic and cubic polynomials with integer coefficients

Integer roots of quadratic and cubic polynomials with integer coefficients Konstantine Zelator Mathematics, Computer Science and Statistics 212 Ben Franklin Hall Bloomsburg University 400 East Second Street

### Factorization in Polynomial Rings

Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18 of Gallian s Contemporary Abstract Algebra. Most of the important

### Breaking The Code. Ryan Lowe. Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and

Breaking The Code Ryan Lowe Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and a minor in Applied Physics. As a sophomore, he took an independent study

### Congruence properties of binary partition functions

Congruence properties of binary partition functions Katherine Anders, Melissa Dennison, Jennifer Weber Lansing and Bruce Reznick Abstract. Let A be a finite subset of N containing 0, and let f(n) denote

### a 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)

ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x

### The Mean Value Theorem

The Mean Value Theorem THEOREM (The Extreme Value Theorem): If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers

### Primality - Factorization

Primality - Factorization Christophe Ritzenthaler November 9, 2009 1 Prime and factorization Definition 1.1. An integer p > 1 is called a prime number (nombre premier) if it has only 1 and p as divisors.

### Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given.

Polynomials (Ch.1) Study Guide by BS, JL, AZ, CC, SH, HL Lagrange Interpolation is a method of fitting an equation to a set of points that functions well when there are few points given. Sasha s method

### Chapter 13: Basic ring theory

Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring

### Kevin James. MTHSC 412 Section 2.4 Prime Factors and Greatest Comm

MTHSC 412 Section 2.4 Prime Factors and Greatest Common Divisor Greatest Common Divisor Definition Suppose that a, b Z. Then we say that d Z is a greatest common divisor (gcd) of a and b if the following

### Non-unique factorization of polynomials over residue class rings of the integers

Comm. Algebra 39(4) 2011, pp 1482 1490 Non-unique factorization of polynomials over residue class rings of the integers Christopher Frei and Sophie Frisch Abstract. We investigate non-unique factorization

### Lecture 6: Finite Fields (PART 3) PART 3: Polynomial Arithmetic. Theoretical Underpinnings of Modern Cryptography

Lecture 6: Finite Fields (PART 3) PART 3: Polynomial Arithmetic Theoretical Underpinnings of Modern Cryptography Lecture Notes on Computer and Network Security by Avi Kak (kak@purdue.edu) January 29, 2015

### SOLUTIONS TO PROBLEM SET 3

SOLUTIONS TO PROBLEM SET 3 MATTI ÅSTRAND The General Cubic Extension Denote L = k(α 1, α 2, α 3 ), F = k(a 1, a 2, a 3 ) and K = F (α 1 ). The polynomial f(x) = x 3 a 1 x 2 + a 2 x a 3 = (x α 1 )(x α 2

### The van Hoeij Algorithm for Factoring Polynomials

The van Hoeij Algorithm for Factoring Polynomials Jürgen Klüners Abstract In this survey we report about a new algorithm for factoring polynomials due to Mark van Hoeij. The main idea is that the combinatorial

### On the generation of elliptic curves with 16 rational torsion points by Pythagorean triples

On the generation of elliptic curves with 16 rational torsion points by Pythagorean triples Brian Hilley Boston College MT695 Honors Seminar March 3, 2006 1 Introduction 1.1 Mazur s Theorem Let C be a

### How To Understand The Theory Of Algebraic Functions

Homework 4 3.4,. Show that x x cos x x holds for x 0. Solution: Since cos x, multiply all three parts by x > 0, we get: x x cos x x, and since x 0 x x 0 ( x ) = 0, then by Sandwich theorem, we get: x 0

### 11 Ideals. 11.1 Revisiting Z

11 Ideals The presentation here is somewhat different than the text. In particular, the sections do not match up. We have seen issues with the failure of unique factorization already, e.g., Z[ 5] = O Q(

### How To Know If A Domain Is Unique In An Octempo (Euclidean) Or Not (Ecl)

Subsets of Euclidean domains possessing a unique division algorithm Andrew D. Lewis 2009/03/16 Abstract Subsets of a Euclidean domain are characterised with the following objectives: (1) ensuring uniqueness

### Algebra 3: algorithms in algebra

Algebra 3: algorithms in algebra Hans Sterk 2003-2004 ii Contents 1 Polynomials, Gröbner bases and Buchberger s algorithm 1 1.1 Introduction............................ 1 1.2 Polynomial rings and systems

### EMBEDDING DEGREE OF HYPERELLIPTIC CURVES WITH COMPLEX MULTIPLICATION

EMBEDDING DEGREE OF HYPERELLIPTIC CURVES WITH COMPLEX MULTIPLICATION CHRISTIAN ROBENHAGEN RAVNSHØJ Abstract. Consider the Jacobian of a genus two curve defined over a finite field and with complex multiplication.

### G = G 0 > G 1 > > G k = {e}

Proposition 49. 1. A group G is nilpotent if and only if G appears as an element of its upper central series. 2. If G is nilpotent, then the upper central series and the lower central series have the same

### An Introduction to the General Number Field Sieve

An Introduction to the General Number Field Sieve Matthew E. Briggs Thesis submitted to the Faculty of the Virginia Polytechnic Institute and State University in partial fulfillment of the requirements

### A New Generic Digital Signature Algorithm

Groups Complex. Cryptol.? (????), 1 16 DOI 10.1515/GCC.????.??? de Gruyter???? A New Generic Digital Signature Algorithm Jennifer Seberry, Vinhbuu To and Dongvu Tonien Abstract. In this paper, we study

### Continued Fractions and the Euclidean Algorithm

Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction

### Some facts about polynomials modulo m (Full proof of the Fingerprinting Theorem)

Some facts about polynomials modulo m (Full proof of the Fingerprinting Theorem) In order to understand the details of the Fingerprinting Theorem on fingerprints of different texts from Chapter 19 of the

### Factoring polynomials over finite fields

Factoring polynomials over finite fields Summary and et questions 12 octobre 2011 1 Finite fields Let p an odd prime and let F p = Z/pZ the (unique up to automorphism) field with p-elements. We want to

### Short Programs for functions on Curves

Short Programs for functions on Curves Victor S. Miller Exploratory Computer Science IBM, Thomas J. Watson Research Center Yorktown Heights, NY 10598 May 6, 1986 Abstract The problem of deducing a function

### RESULTANT AND DISCRIMINANT OF POLYNOMIALS

RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results

### 3.6 The Real Zeros of a Polynomial Function

SECTION 3.6 The Real Zeros of a Polynomial Function 219 3.6 The Real Zeros of a Polynomial Function PREPARING FOR THIS SECTION Before getting started, review the following: Classification of Numbers (Appendix,

### p e i 1 [p e i i ) = i=1

Homework 1 Solutions - Sri Raga Velagapudi Algebra Section 1. Show that if n Z then for every integer a with gcd(a, n) = 1, there exists a unique x mod n such that ax = 1 mod n. By the definition of gcd,

### A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number

Number Fields Introduction A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number field K = Q(α) for some α K. The minimal polynomial Let K be a number field and

### MATH 289 PROBLEM SET 4: NUMBER THEORY

MATH 289 PROBLEM SET 4: NUMBER THEORY 1. The greatest common divisor If d and n are integers, then we say that d divides n if and only if there exists an integer q such that n = qd. Notice that if d divides

### Collinear Points in Permutations

Collinear Points in Permutations Joshua N. Cooper Courant Institute of Mathematics New York University, New York, NY József Solymosi Department of Mathematics University of British Columbia, Vancouver,

### GREATEST COMMON DIVISOR

DEFINITION: GREATEST COMMON DIVISOR The greatest common divisor (gcd) of a and b, denoted by (a, b), is the largest common divisor of integers a and b. THEOREM: If a and b are nonzero integers, then their

### Real Roots of Univariate Polynomials with Real Coefficients

Real Roots of Univariate Polynomials with Real Coefficients mostly written by Christina Hewitt March 22, 2012 1 Introduction Polynomial equations are used throughout mathematics. When solving polynomials

### HOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)!

Math 7 Fall 205 HOMEWORK 5 SOLUTIONS Problem. 2008 B2 Let F 0 x = ln x. For n 0 and x > 0, let F n+ x = 0 F ntdt. Evaluate n!f n lim n ln n. By directly computing F n x for small n s, we obtain the following

### k, then n = p2α 1 1 pα k

Powers of Integers An integer n is a perfect square if n = m for some integer m. Taking into account the prime factorization, if m = p α 1 1 pα k k, then n = pα 1 1 p α k k. That is, n is a perfect square

### Chapter 7: Products and quotients

Chapter 7: Products and quotients Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 7: Products

### Applications of Fermat s Little Theorem and Congruences

Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4

### Introduction to Algebraic Coding Theory

Introduction to Algebraic Coding Theory Supplementary material for Math 336 Cornell University Sarah A. Spence Contents 1 Introduction 1 2 Basics 2 2.1 Important code parameters..................... 4

### On the irreducibility of certain polynomials with coefficients as products of terms in an arithmetic progression

On the irreducibility of certain polynomials with coefficients as products of terms in an arithmetic progression Carrie E. Finch and N. Saradha Abstract We prove the irreducibility of ceratin polynomials

### Notes 11: List Decoding Folded Reed-Solomon Codes

Introduction to Coding Theory CMU: Spring 2010 Notes 11: List Decoding Folded Reed-Solomon Codes April 2010 Lecturer: Venkatesan Guruswami Scribe: Venkatesan Guruswami At the end of the previous notes,